Show that the derivative of a function is not continuous

Question

$$g(x)=\begin{cases} x+2x^2\sin\left(\frac{1}{x}\right)&\text{ if }x\neq0\\\ 0&\text{ if }x=0 \end{cases}$$ Show that there is a sequence $\{x_n\}$ with $\{x_n\} \to 0$ as $n$ approaches infinity, such that $g'(x_n)=0\ \forall n$ but $g'(0) \ne 0$.

I calculated $$g'(x)=4x\sin\left(\frac{1}{x}\right)-2\cos\left(\frac{1}{x}\right)+1 ,\text{ if }x\neq0$$

I tried to construct a sequence where $\{x_n\}=\dfrac 1 {2\pi n}$ so that $g'(x_n)=0$, but I'm not sure if this is the right way to prove the question. If this is not right, how can I go about showing it?

Answer 1

The first summand of your $g'$, $$4x\sin\left(\frac{1}{x}\right)$$ converges to $0$ if $x$ converges to $0$, because $|4\sin\left(\frac{1}{x}\right)|\le 4$.

$$\cos(\phi)=\frac{1}{2}, \text{iff} \;\phi=\pm\frac{\pi}{3}\pm 2k\pi$$

so choose $$\frac{1}{x}=\pm\frac{\pi}{3}\pm 2k\pi$$ e.g. $$x_n=\frac{1}{\frac{\pi}{3}+ 2n\pi}$$

But in a similar way you can construct sequences $x_n$ that converge to a value different to $0$.

$$g'(0)=\lim_{x \to 0}\left(\frac{g(x)-g(0)}{x-0}\right)=\lim_{x \to 0}\left(1-x\sin\left(\frac{1}{x}\right)\right)=1$$ So the constructed sequence shows that the derivate $g'$ of $g$ is not continous at $x=0$.